Using P-values to make conclusions (article) | Khan Academy (2024)

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• Stan

  6 years agoPosted 6 years ago. Direct link to Stan's post “Could any one explain how...”

  Could any one explain how to get the p-value in the second example?

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  • Saxon Knight

    5 years agoPosted 5 years ago. Direct link to Saxon Knight's post “Sure!The p-value is the...”

    

    

    

    Sure!

    The p-value is the probability of a statistic at least as deviant as ours occurring under the assumption that the null hypothesis is true.

    Under that assumption, and noting also that we are given that the population is normally distributed (or that we took a sample size of at least 30 [by the Central Limit Theorem]), we can treat the sampling distribution of the sample mean as a normal distribution.

    So now, we can use the normal cumulative density function or a z-table to find this probability. (We could also use a t-table, but it is allowable to just use a z table since our sample size is larger than 30)

    To use a z-table, we'll need to find the appropriate z-score first.

    Since the answer to what we are asking comes from the sampling distribution of the sample mean, we would find the appropriate standard deviation to use by dividing the population standard deviation by the square root of the sample size (since the variance of the sampling distribution is the population variance divided by the sample size, and the standard deviation is the square root of the variance).

    That would give us a standard deviation for the sampling distribution of the sample mean.

    I say would, because unfortunately, we don’t always know the population standard deviation, and so (as it seems they did here, despite knowing the population standard deviation), we are using the sample standard deviation in its place to find an estimate of the standard deviation for the sampling distribution of the sample mean, which is also known as the standard error of the mean.

    In our example, the standard error of the mean therefore has a value of 0.12 / 50^0.5, or approximately 0.01697.

    Taking the difference between our sample mean and the population mean and dividing it by the standard error gives us our z-score (number of standard errors our sample mean is away from the population mean), which is approximately (7.36 - 7.4) / 0.01697 or -2.36.

    Since the alternative hypothesis is not specific about the population mean being either greater than or less than the value in the null hypothesis, we have to consider both tails of the distribution, but by symmetry of the standard normal distribution, we can accomplish this by simply doubling the value we get from using our obtained z-score with a z-table.

    The value given by a z-table using a z-score of -2.36 is 0.0091, which, when doubled, is 0.0182 or approximately 0.02.

    This (or other videos before it in that section) might also help (it comes later in this unit): https://www.khanacademy.org/math/statistics-probability/significance-tests-one-sample/tests-about-population-mean/v/calculating-p-value-from-t-statistic

    :)

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    (57 votes)
• G.Gulzt

  6 years agoPosted 6 years ago. Direct link to G.Gulzt's post “I don't understand the p-...”

  I don't understand the p-value in example 1

  Isn't the calculation: binomial(60,20) * 0.75^40 * 0.25^20 = 0.0383?
  The problem states it is "0.068". Is this p-value wrong or did I make a mistake in my calculation?

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  (6 votes)

  • Mahmood Salah

    See Also

    Den P Wert berechnen: 7 Schritte (mit Bildern) – wikiHowHow To Find P Hat: Understanding, Calculating, And Interpreting P Hat – ProLearnPointDen p-Wert in der Statistik verstehen und interpretierenp Wert: einfache Erklärung & Berechnung

    6 years agoPosted 6 years ago. Direct link to Mahmood Salah's post “p-value = P(p(x) >= 20/60...”

    p-value = P(p(x) >= 20/60 given that the actual proportion is 0.25)
    So, You need to calculate:
    binomial(60,20) * 0.75^40 * 0.25^20 (probability of 20 subject that identified the bottled water )
    +
    binomial(60,21) * 0.75^39 * 0.25^21 (probability of 21 subject that identified the bottled water )
    +
    binomial(60,22) * 0.75^38 * 0.25^22
    +
    binomial(60,23) * 0.75^37 * 0.25^23
    :
    :
    binomial(60,60) * 0.75^0 * 0.25^60 (probability of all the 60 subjects identified the bottled water )

    I guess !

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    (8 votes)
• Nishay

  6 years agoPosted 6 years ago. Direct link to Nishay's post “How do you decide what Si...”

  How do you decide what Significance level you should set??

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  (1 vote)

  • Ian Pulizzotto

    6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “A significance level of 0...”

    

    A significance level of 0.05 (i.e. 5%) is commonly used, but sometimes other significance levels are used.

    Note that the significance level is the probability of a Type 1 error (rejecting a true null hypothesis). Everything else being equal, decreasing the significance level (probability of a Type 1 error) increases the probability of a Type 2 error (failing to reject a false null hypothesis), and vice versa.

    So the statistician has to weigh the cost of a Type 1 error (rejecting a true null hypothesis) versus the cost of a Type 2 error (failing to reject a false null hypothesis) in the real-world situation. If the statistician is especially concerned about the cost of a Type 1 error, then he/she will use a significance level that is less than 0.05. However, if instead the statistician is especially concerned about the cost of a Type 2 error, then he/she will use a significance level that is greater than 0.05.

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    (12 votes)
• Mohammad Yasser

  5 years agoPosted 5 years ago. Direct link to Mohammad Yasser's post “As far as I understand, r...”

  As far as I understand, rejecting H0 doesn't mean accepting Ha in all cases. Rejecting H0 only implies accepting Ha iff both are complements to each other, i.e. exactly one of them must be true. E.g. if H0 says x = 5, and Ha says x > 5, then maybe both are wrong and the truth is x < 5. This will be so weird though because the truth is expected to be either H0 or Ha, but I think it's theoretically possible to happen.

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  • Nick Barnes

    a year agoPosted a year ago. Direct link to Nick Barnes's post “You're confounding the tr...”

    You're confounding the truthfulness of H0 with the acceptability of Ha. In your example, not accepting Ha says we will not accept that x > 5, in other words x = 5 or x < 5. Not accepting Ha does not report on the truth that x < 5, it still allows the possibility that x = 5 - that is H0 is not rejected. It's very tempting to say H0 is "rejected" because x = 5 is a false statement. The key is to clarify what is meant by "reject". The statistics notion of reject is not based on whether the hypothesis is a true or false statement but on if it is rejected by the acceptability criteria of Ha.

    From that perspective verify these statements (the logic flows from one to the next): If you do not accept Ha, then you do not reject H0. The only way you can reject H0 is by accepting Ha. It doesn't make sense to both reject H0 and not accept Ha.

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    (1 vote)
• VVCephei

  6 years agoPosted 6 years ago. Direct link to VVCephei's post “In the first problem, is ...”

  See Also

  Sampling Distribution of the Sample Proportion, p-hat

  In the first problem, is 0.068 the correct p-value? Assuming that the null hypothesis is true, and p = 0.25, the sampling distribution of sample proportion with n = 60 should be approximately normal, with a mean = p = 0.25 and standard deviation of √((p·(1-p))/n) ≈ 0.056. So a sample with p-hat = 0.3 should only have a z-score ≈ 0.89, and there should be ≈ 0.187 probability of getting a sample with p-hat ≥ 0.3. Or am i missing something?

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  (2 votes)

  • Ian Pulizzotto

    6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “You generally had the rig...”

    You generally had the right idea for calculating the p-value. Note that the p-hat value is not 0.3, but rather 20/60 = 1/3 = 0.3333... (perhaps you did not consider the bar on top of the decimal digit 3). So the z-score is about 1.49 instead of 0.89. The probability of equaling or exceeding a z-score of 1.49 is about 0.068.

    Have a blessed, wonderful day!

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• JorgeMercedes

  6 years agoPosted 6 years ago. Direct link to JorgeMercedes's post “Please please please show...”

  Please please please show us how those P Values come about. We are very troubled.

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  • mzg23patel

    6 years agoPosted 6 years ago. Direct link to mzg23patel's post “the p values are generall...”

    the p values are generally given whenever such problems are asked, i think calculating p values is a completely unrelated concept here so it is not taught

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    (5 votes)
• A_meginniss

  a year agoPosted a year ago. Direct link to A_meginniss's post “what is the equation to c...”

  what is the equation to calculate the p value

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  (3 votes)

  • daniella

    3 months agoPosted 3 months ago. Direct link to daniella's post “The equation to calculate...”

    The equation to calculate the p-value depends on the specific hypothesis test being performed. For example:
    In a z-test for a population mean, the p-value can be calculated using the standard normal distribution tables or software functions.
    In a t-test for a population mean, the p-value is typically calculated using the t-distribution tables or software functions.
    In a chi-square test for independence, the p-value is calculated based on the chi-square distribution.
    Each test has its own formula for calculating the p-value based on the observed sample data and the assumptions of the test.

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    (1 vote)
• Junsang

  5 years agoPosted 5 years ago. Direct link to Junsang's post “ur... are we going to be ...”

  ur... are we going to be told how to calculate this P-value?

  I'm confused on what it actually is...

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  (2 votes)

  • Jerry Nilsson

    5 years agoPosted 5 years ago. Direct link to Jerry Nilsson's post “Yes, there are lessons on...”

    Yes, there are lessons on how to calculate the p-value, which is the probability that the assumed population parameter is true, based on a sample statistic of that parameter.

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    (2 votes)
• ricardoadam_

  4 years agoPosted 4 years ago. Direct link to ricardoadam_'s post “First problem, question B...”

  First problem, question B, remark for answer C
  "There wasn't enough evidence to reject H0 at this significance level, but that doesn't mean we should accept H0. This experiment didn't attempt to collect evidence in support of H0."

  What would be like an experiment that would collect evidence in support of H0?

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  • Bryan

    4 years agoPosted 4 years ago. Direct link to Bryan's post “This experiment just assu...”

    This experiment just assumed Ho was true; if p-value was below our sig level, then our assumption of Ho could be rejected, since it's unlikely we'd get such a deviant (or more deviant) sample proportion if Ho was true.
    If p-value was above our sig level, it tells us that Ha can be rejected, since it's likely enough to get a sample proportion of 0.333333333etc or more assuming Ho; there is no need for Ha to be true (no need for pop proportion to be higher).

    But Ha being rejected doesn't prove Ho (pop proportion = 0.25).

    For example, our hypothesis could be
    Ho: p = 0.245
    Ha: p > 0.245

    And then with a p^ of 0.33333etc, we would have a p-value of around 0.056, which still above our sig level, meaning that we reject our Ha, p > 0.245.
    This would be a contradiction if our first Ho was proven, but it wasn't, so it's not a contradiction.
    Notice how rejecting p > 0.245 doesn't conflict with rejecting p > 0.25, since we never said p had to be in between 0.245 and 0.25.

    You could maybe use the law of large numbers and coerce millions of people into guessing the water in your cups, and see if that proportion is really close to 0.25 to possibly prove that p = 0.25. There's probably a better experiment, but I'm not too experienced in thinking of them :P

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    (1 vote)
• brittshi000

  6 months agoPosted 6 months ago. Direct link to brittshi000's post “Thank you for the great q...”

  Thank you for the great questions. They helped me so much to prepare for my test.

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  (2 votes)